CHAPTER 12
1
Chemical Bonding
CHAPTER ANSWERS
1.
bojid
2
bond energy
3.
ionic
4.
covalent
5.
In Cl
2 the bonding is pure covalent with the electron pair shared equally between the two chlorine
atoms. In HCI there is a shared pair of electrons, but the shared pair is drawn more closely o the
chlorine atom, and this makes the bond polar.
6.
In 112 and HF, the bonding is covalent in nature with an electron pair being shared between the
atoms. In 112, the two atoms are identical (the sharing is equal). in HF, the two atoms are different
(the sharing is unequal) and as a result the bond is poLar. Both of these are in marked contrast to
the situation in NaF: NaF is an ionic compound—an electron has been completely transferred
ITom sodium to fluorine, producing separate ions.
7.
elect ronegativity
A bond is polar if the centers of positive and negative charge do not coincide at the same point.
The bond has a negative end and a positive end. Polar bonds will exist in any molecule with
nonidentical bonded atoms (although the molecule as a whole may not be polar if the bond dipoles
cancel each other). Two simple examples are HF and HCl; in both cases, the negative center of
charge is closer to the halogen atom.
.
9.
from top to bottom: covalent, ionic, polar covalent
10.
The level of polarity in a polar covalent bond is determined by the difference in etectronegativity
of the atoms in the bond.
II.
In general, an element farther to the right in a given period or an element closer to the top of a
given group is more electronegative. The elements below are ranked in order of increasing
electronegativity.
a.
Li<C<F
b.
1<C1<F
c.
Cs<Rb<Li
a.
I is most electronegative, Rb is least electronegative
b.
Mg is most eIectronegatie, Ca and Sr have similar electronegativities
c.
Br is most electronegative, K is least electronegatk.e
12.
pr gvl
Hogh.tn Mifflin Compeny All rgsts reserved
248
13.
14.
15.
16.
17.
18.
19.
Chapter 12: Chemical Bonding
Generally, covalent bonds between atoms of difJrent elements are polar.
a.
ionic
b.
ionic
c.
polar covalent
d.
polar covalent
Generally, covalent bonds between atoms of different elements are polar.
a.
ionic
b.
polar covalent
c.
covalent
For a bond to be polar covalent, the atoms involved in the bond must have different
electronegativities (must be of different elements).
a.
nonpolar covalent (atoms of the same element)
b.
nonpolar covalent (atoms of the same element)
c.
nonpolar covalent (atoms of the same element)
d.
polar covalent (atoms of different elements)
For a bond to be polar covalent, the atoms involved in the bond must have different
electronegativities (must be of different elements).
a.
nonpolar covalent (atoms of the same element)
b.
nonpolar covalent (atoms of the same element)
c.
polar covalent (atoms of different elements)
d.
polar covalent (atoms of different elements)
The degree of polarity of a polar covalent bond is indicated by the magnitude of the difference in
electronegativities of the elements involved; the larger the difference in electronegativity, the
more polar the bond. Electronegativity differences are given in parentheses below:
a.
H—F (1.9); H—Cl (0.9); the H—F bond is more polar
b.
H—Cl (0.9); H—I (0.4); the H—Cl bond is more polar
c.
H—Br (0.7); H—Cl (0.9); the H—Cl bond is more polar
d.
H—I (0.4); H—Br (0.7); the H—Br bond is more polar
The atom with the larger electronegativity will be more negative relative to the other atom.
a.
F
b.
neither (similar electronegativity)
c.
Cl
d.
(‘I
The larger the diflèrence in electronegativity between h o atoms. the more polar will be the bond
between those atoms. Elecironegativity differences are given in parentheses below:
a.
H—S (0.4); H—F (1 .9): the H—F bond is more polar
Copy ç:
.‘
-DLghor
r Corpar A
:s resevrJ
Chapter 12: Chemical Bonding
h.
249
0—S (I .0); 0 F (0.5): the 0 S bond is more polar
N-—S (0.5); N (1(0); the N -S bond is more polar
ti.
20.
21.
C—S (0); C—-Cl (0.5); the C--Cl bond is more polar
The greater the electronegativity dilYcience between two atoms. the more ionic will be the bond
between those two atoms.
a.
Ca—Cl
b.
Ba-Cl
c.
Fe—I
d.
Be—F
A dipole moment is an electrical elYect that occurs in a molecule that has separate centers of
positive and negative charge. The simplest examples ol’molecules with dipole moments ould be
diatomic molecules involving two diflrent elements. For example:
ö+N->Oö6+CL—F 8—
8+ Br—Cl 8—
22.
The presence of strong bond dipoles and a large overall dipole moment in water make it a polar
substance overall. Among the properties of’ water dependent on its dipole moment are its freezing
point, melting point, vapor pressure, and its ability to dissolve many substances.
23.
In a diatomic molecule containing two different elements, the more electronegative atom will be
the negative end of the molecule, and the less electronegative atom will be the positive end.
24.
25.
a.
chlorine
b.
oxygen
c.
fluorine
In a diatomic moiecu[e containing two ditèrent elements, the more electronegative atom will be
the negative end of the moicue, aa the less electronegative atom will be the positive end.
a.
H
b.
CI
c.
I
In the figures, the arrow points toward the more electronegative atom.
a.
b.
6-t-Si---C6--
c.
d.
26.
3--3-->C&-
.s t.:nw the :.‘c:: .ec:rorga:v a:oi.
in the grec
a.
f;y
n-o or
.
c:
-
.
250
Chapter 12 Chemical BondIng
&fP—*O&-
b.
..
C,
P and H have similar electronegativities.
d.
27.
In the figures, the arrow points toward the more electronegative atom.
a.
28.
b.
P—H The atoms have very nearly the same electronegativity, so there is a very small, if any,
dipole moment.
C.
il-t—S3.
d.
+l1—*CH
In the
figures,
the armw points toward the more electronegative atom.
a,
b.
+N—,O
c.
d.
&tC—N
29.
When forming ionic bonds, the element forming the positive ion loses enough electrons to have
the same configuration as the previous noble gas. The element ftwming the negative ion gains
enough electrons to have the same contiguration as the next noble gas. When forming covalent
compounds, electrons are shared in such a way that as many atoms as possible in the molecule
have a coiitiguraiion analogous to a noble gas.
30.
preceding
31.
gaining
32.
Atoms in covalent molecules gain a conhguration like that of a noble gas by sharing one or more
pairs of electrons between atoms. Such shared pairs of electrons “belong” to each of the atoms of
the bond at the same time. In ionic bonding. one atom completely gives over one or more
electroiis to another atom, and the resulting ions behave independently of one another.
33
a.
Na
ls 2s 2p I, 3s I
Na
I.% 2.2 2i/
Ne has the same configuration as Na.
I,.
I
F
c.
I s 2.s’
2
2)1 32
3p’ 4,v
2 3d 4p 5.s
2
4dH
5p
5
2 2s
is
2 2p’ 3s 3p’ 4 3d 4p” 5s
2 4d
1 5po1
Xe has the same configuration as I
Ca
I 2 252 2? 32 3 42
’
2
Ca
2 252 2
Is
3,2
6
3p
Ar has the same configuration as Ca
2
Copyright
Houghton Miflin Company All rights reserved
Chapter 12: Chemical Bonding
N
d.
251
2 252 2p
Is
3
%2
2 2i’
2.v
3
No has the same con tiguration as N
L
F
Is 2s’2p’
F
2 2p
2 2s
Is
Ne has the same con tiguration as F.
34.
36.
a.
Br, Kr (Br has one electron less than Kr.)
h.
Cs, Xe (Cs has one electron more than Xe.)
c.
, Ar (P has three fewer electrons than Ar.)
3
P
d.
, Ar (S has two fewer electrons than Ar.)
2
S
a.
2 (Mg has two electrons more than the noble gas Ne.)
Mg
b.
3 (Al has three electrons more than the noble gas Ne.)
Al
c.
F (I has one electron less than the noble gas Xe.)
d.
Ca (Ca has two electrons more than the noble gas Ar.)
Atoms or ions with the same number of electrons are said to be isoelectronic.
a.
3
F, 02, N
b.
3
Cl, S
,P
2
c.
N
3
,
2
F, O
d.
3
Br, S02 As
a.
A1 Al has three electrons more than a noble gas; S has two fewer electrons than a noble
:
3
S
2
gas.
h.
RaO: Ra has two electrons more than a noble gas; 0 has two fewer electrons than a noble
gas.
c.
: Ca has two electrons more than a noble gas; F has one electron less than a noble gas.
2
CaF
ci.
CsN; Cs has one electron more than a noble gas; N has three fewer electrons than a noble
gas.
e.
RbP: Rb has one electron more than a noble gas; P has three fewer electrons than a noble
gas.
a.
\aS: Na has one electron more than a noble gas; S has two fewer electrons than a noble
gas.
37.
3.
.
r ç’
BaSe: Ba has wo electrons more than a noble gas; Se has two fewer electrons than a noble
gas.
iogh:oi ‘ñ ft r Cor’pa.’y A. -ighfs -eseved.
252
Chapter 12: Chemical Bonding
c.
: Mg has two electrons more than a noble gas; Br has one electron less than a noble
2
MgBr
gas.
d.
LiN: Li has one electron more than a noble gas; N has three fewer electrons than a noble
gas.
e.
1(1-I: K has one electron more than a noble gas; H has one electron less than a noble gas.
a.
, [Xe]; S
24
Ba
, [Ar]
2
b.
, [Kr]; F, [Ne]
2
Sr
c.
, [Ne]; O2, [Ne]
2
Mg
d.
, [Ne]; S
3
Al
, [Ar]
2
a.
, [Kr]; 02_, [Ne)
2
Sr
b.
, [Ar]; H, [He]
2
Ca
c.
K, [Ar]; P
, [Ar]
3
d.
Ba2+ [Xe]; Se 2— [Kr]
39.
40.
,
,
41.
The formula of an ionic compound represents only the smallest whole number ratio of positive
and negative ions present (the empirical formula).
42.
An ionic solid such as NaCl consists of an array of alternating positively and negatively charged
ions; that is, each positive ion has as its nearest neighbors a group of negative ions, and each
negative ion has a group of positive ions surrounding it. In most ionic solids, the ions are packed
as tightly as possible.
43.
Positive ions are always smaller than the atoms from which they are formed because, in forming
the ion, the valence electron shell (or part of it) is “removed” from the atom.
44.
In forming an anion, an atom gains additional electrons in its outermost (valence) shell.
Additional electrons in the valence shell increase the repulsive forces between electrons, so the
outermost shell becomes larger to accommodate this.
45.
Relative ionic sizes are given in Figure 12.9. Within a given horizontal row of the periodic chart,
negative ions tend to be larger than positive ions because the negative ions contain a larger
number of electrons in the valence shell. Within a vertical group of the periodic table, ionic size
increases from top to bottom. In general, positive ions are smaller than the atoms from which they
come, whereas negative ions are larger than the atoms from which they come,
a.
H
b.
N
c.
Al 3-
d.
F
Copyrc
hokgtc1
.
‘ese’vec
Chapter 12: ChemIcal BondIng
46.
I
47.
414
253
Relative ionic sizes are given in Figure 12.9. Within a given horizontal row of the periodic chart,
negative ions tend to be larger than positive ions because the negative ions contain a larger
number of electrons in the valence shell. Within a vertical group of the periodic table, ionic size
increass from top to bottom. In general, positive ions are smaller than the atoms from which they
come, whereas negative itins are larger than the atoms from which they come.
a.
V
b.
Cl
c.
Ca
d.
1
Relative ionic sizes are given in Figure 12.9. Within a given horizontal row of the periodic chart,
negative ions tend to be larger than positive ions because the negative ions contain a larger
number of electrons in the valence shell. Within a vertical group of the periodic table, ionic size
increases from top to bottom. In general, positive ions are smaller than the atoms from which they
come, whereas negative ions are larger than the atoms from which they come.
a
Fe
b
Cl
c
Al
Relative ionic. sizes are given in Figure 12 9 Wiihm a given horizontal row of the periodic uhart,
negative ions tend to be larger than positive ions because the negative ions contain a larger
number of electrons in the valence shell Within a vertical group of the periodic table ionic size
incieases from top to bottom In general, positive ions are smaller than the atoms from which they
tome, whereas negative ions are larger than the atoms from which they come
a.
I
b.
F
c.
F
49.
Valence electrons are those found in the outermost principal energy level of the atom. The valence
electrons effectively represent the outside edge of the atom and are the electrons most influenced
by the electrons of another atom.
50.
When atoms form covalent bonds, they try to attain a valence electronic configuration similar to
that of the tbllowing noble gas element. When the elements in the first few horizontal rows of the
periodic table form covalent bonds, they will attempt to gain configurations similar to the noble
gases helium (two valence electrons, duet rule), and neon and argon (eight valence electrons, octet
rule).
5 I.
noble gas electronic configuration
52.
These elements attain a total of eight valence electrons, making the valence electron
configurations similar to those of the noble gases Ne and Ar.
53.
When two atoms in a molecule are connected by a double bond, the atoms share two pairs of
electrons (four electrons) in completing their outermost shells. A simple molecule containing a
4 (l-1
C::CF-b).
1
double bond is ethene (ethylene), CH
54.
When two atoms in a molecule are connected by a triple bond, the atoms share three pairs of
electrons (six electrons) in completing their outermost shells. A simple molecule containing a
triple bond is acetylene, 2
Fl (l-1:C:::C:H).
C
Copyngt
Hoighton MffHn Company. All r,grits reserved.
254
Chapter 12: Chemical Bonding
55.
a.
:1:
b.
AI
C.
Xe:
d.
Sr
56.
a
Rb•
b.
.-,
C.
4
B.
a.
e.
4
‘p.
1.:
At
57,
a.
each N provIdes 5; 0 provides 6; total valence electrons
b.
each B provides 3; each H provides 1: total valence electrons
=
12
c.
each C provides 4; each H provides 1; total valence electrons
=
20
d.
N provides 5; each CL provides 7; total valence electrons
a.
phosphorus provides 5; each hydrogen provides 1; total valence electrons
b.
each N provides 5; oxygen provides 6; total valence electrons
c.
each carbon provides 4; each hydrogen provides 1; total valence electrons
=
16
26
58.
=
8
16
=
26
Copyright © Hoighton Miffn Company AB rights reserved
Chapter 12: Chemical Bonding
255
d.
each carbon provides 4; each hydrogen prçvides 1; bromine provides 7; total valence
electrons = 26
a.
Nitrogen provides 5 valence electrons, each bromine provides 7, total ‘valence electrons =
26.
59.
Br—NBr
-
I
Br
b.
Hydrogen provides 1 valence electron; fluorine provides 7; total valence electrons
8.
H—F:
c.
carbon provides 4 valence electrons; each bromine provides 7; total valence electrons
=
32.
J3r:
Br—C—Br
Br
d
Each carbon provides 4 valence electrons, each hydrogen provides 1, total valence eIeLtrons
=10
H
H
/
/
\
H
H
60.
a
Each hydrogen provides 1 valence electron, total valence electrons
b.
H—K
Hydrogen provides I valence electron; chlorine provides 7 valence electrons; total valence
electrons 8
2
H—Cl
c
Carbon provides 4 valence electrons each fluorine provides 7 valence electrons, total
valence electrons 32
F-b-F
F
Copyright
Houghton MHThn Company. AI rights reserved.
256
Chapter 12: Chemical Bonding
d.
Each carbon provides 4 valence electrons; each fluorine provides 7 valence electrons; total
valence electrons = 50
:F:
:F:
I..
I
I
I
F—C—C——F:
:F:
:F
1
61.
a,
Each C provides 4 valence electrons. Each H provides I valence electron. Total valence
electrons 14
H
II
I
Fl—C
I
1-1
h.
C
—
—
H
H
N provides 5 valence electrons. Each F provides 7 valence electrons. Total valence electrons
26
:F—N—F:
F:
e.
Each C provides 4 valence electrons. Each H provides I valence electron. Total valence
electrons 26
1-I
H
H
11
IL-C-C-C-C-H
liii
d.
H Fl [1 H
Si provides 4 valence electrons. Each Cl provides 7 valence electrons. Total valence
electrons 32
4
:Cl—Sj—Cl:
I
•‘
:Cl:
62,
a.
Each H provides I valence electron. S provides 6 valence electrons. Total valence electrons
=8
H—S-—H or sometimes drawn as follows to indicate the shape H—S:
H
Copyright
Houghion Miffiin Company Al rights rasered.
Chapter 12: Chemical Bonding
b.
257
Si provides 4 valence electrons. Each F provides 7 valence electrons. Total valence electrons
=32
:F—Sj——F:
c.
d.
Each C provides 4 valence electrons. Each H provides I valence electron. Total valence
electrons = 12
H
11
H H
or
I I
H—CC—H
H
H
Each C provides 4 valeiee electrons. Each H provides I valence electron. Total valence
electrons = 20
H
H
H
[I
II
ii
I I I
K—C—C—C—H
I I I
63.
a.
no resonance
:NN—O:
b.
The fact that there is an odd number of electrons (17) guarantees resonance structures. Here
are two possible structures.
c.
Resonance is possible in the nitrate ion. Here are three structures showing only the bonding
electrons.
0
0
0
II
I
I
64.
a.
N provides 5 valence electrons. Each 0 provides 6 valence electrons. Total valence electrons
= 17 (an odd number). Note that there are several Lewis structures possible because of the
odd number of electrons (the unpaired electron can be located on different atoms in the
Lewis structure).
0N—O
Gopyr ght © Houghton Mifflin Company All oghts re9erved.
0—N0
0—N0
258
Chapter 12: ChemIcal Bonding
b.
Each H provides 1 valence electron. S provides 6 valence electrons. Each 0 provides 6
valence electrons. Total valence electrons = 32
::
H—ØS—Q—H
:0:
c.
Each N provides 5 valence electrons. Each 0 provides 6 valence electrons. Total valence
electrons 34
:0: :0:
:0: :0;
:O—N—N—O:
65.
a.
S provides 6 valence electrons. Each 0 provides 6 valence electrons. The 2— charge means
two additional valence electrons. Total valence electrons = 32
1 2-
b.
P provides 5 valence electrons. Each 0 provides 6 valence electrons. The 3— charge means
three additional valence electrons. Total valence electrons 32.
1 3-
c.
.4
S provides 6 valence electrons. Each 0 provides 6 valence electrons. The 2— charge means
two additional valence electrons. Total valence electrons 26
:O:
;O—S—O:
66.
a.
4
Cl provides 7 valence electrons. Each 0 provides 6 valence electrons. The 1— charge means
1 additional electron. Total valence electrons = 26
:0:
:O—Cl—O:
11
Copyright © Houghton Mflhin Company All rights reserved
Chapter i2 Chemical Bonding
259
b.
Each 0 provides 6 valence electrons. The 2—charge means two additional valence electrons.
Total valence electrons = 14
c,
Each C provides 4 valence electrons. Each H provides I valence electron. Each 0 provides 6
The 1— charge means I additional valence electron, Total valence
valence electrons.
electrons = 24
1-
[10
I
II
H—C—C—0:
L
H
67.
a.
N provides 5 valence electrons. Each 0 provides 6 valence electrons. The 1— charge means
one additional valence electron. Total valence electrons = 18
[ors.i_o1
b
1
[O—No]’
[1 provides I valence eleLtron C provides 4 valence electrons Each 0 provides 6 valence
electrons. The 1— charge means one additional valence electron. Total valence electrons 24
:0:
:O:
H—O—C=O
c.
0 provides 6 valence electrons. H provides I valence electron. The 1— charge means one
additional valence electron. Total valence electrons 8
[:—ii
Ii
68.
a.
I
1-I provides I valence electron. P provides 5 valence electrons. Each 0 provides 6 valence
electrons. The 2— charge means two additional valence electrons. Total valence electrons =
32.
:ö:
2-
H—0—P—0
Copyngrit © Houghton Mifflin Company. All rights reserved.
260
Chapter 12: Chemical Bonding
b.
Each H provides 1 valence electron. P provides 5 valence electrons. Each 0 provides 6
valence electrons. The I charge means one additional valence electron. Total valence
electrons = 32
—
:0:
H——P—Q—F1
:0:
c.
P provides 5 valence electrons. Each 0 provides 6 valence electrons. The 3— charge means
three additional valence electrons. Total valence electrons 32
1 3-
69.
The geometric structure of the water molecule is bent (or V-shaped). There are four pairs of
valence electrons on the oxygen atom of water (two pairs are bonding pairs; two pairs are nonbonding lone pairs). The H.—O—H bond angle in water is approximately 106°.
70.
The geometric structure of NH
3 is that of a trigonal pyramid. The nitrogen atom of NH
1 is
surrounded by four electron pairs (three are bonding: one is a lone pair). The 1-1--N 1-1 bond angle
is somewhat less than 109,5° (due to the presence of the lone pair).
71.
BF is described as having a trigonal planar geometric structure, The boron atom of HF
3
3 is
surrounded by only three pairs of valence electrons, The F—B—F bond angle in HF
3 is 120°.
72.
The geometric structure of SiF
4 is that of a tetrahedron. The silicon atom of SiP
4 is surrounded by
four bonding electron pairs. The F—Si--F bond angle is the characteristic angle of the tetrahedron,
109.5°.
73.
The geometric structure of a molecule plays a very important part in its chemistry. For biological
molecules, a slight change in the geometric structure of the molecule can completely destroy the
molecul&s usefulness to a cell or can cause a destructive change in the cell.
74.
The general molecular structure of a molecule is determined by (I) how many electron pairs
surround the central atom in the molecule and (2) which of those electron pairs are used for
bonding to the other atoms of the molecule. Nonbonding electron pairs on the central atom do,
however cause minor changes in the bond angles compared to the Ideal regular geometric
structure.
75.
For a given atom, the positions of the other atoms bonded to it are determined by maximizing the
angular separation of the valence electron pairs on the given atom to minimize the electron
repulsions.
76.
You will remember from high school geometry that two points in space are all that is needed to
define a straight line. A diatomic molecule represents two points (the nuclei of the atoms) in
space.
Copyrlht © Hrnighton Mifflin Company Alt rights resetved
Chapter 12: ChemicaL BandIng
261
77.
One of the valence electron pairs of the ammonia molecule is a lone pair with no hydrogen atom
attached. This lone pair is part of the nitrogen atom and does not enter the description of the
molecule’s overall shape (aside from influencing the H—N—H bond angles between the remaining
valence electron pairs).
78.
3 there are only three
, the nitrogen atom ha four pairs of valence electrons whereas in BF
3
in NP
pair on nitrogen in
electron
boron
The
nonbonding
atom.
the
around
electrons
valence
of
pairs
atom.
the
the
plane
N
of
atoms
of
out
3 pushes the three F
NF
79.
If you draw the Lewis structures for these molecules, you will see that each of the indicated atoms
is surrounded by fhur pairs of electrons with a tetrahedral orientation of the electron pairs (the
S is distorted because some electron pairs on the S atom are bonding and some
2
arrangement in H
are nonbonding).
80.
a.
four electron pairs arranged in a tetrahedral arrangement with some distortion due to the
nonbonding pair
b.
four electron pairs in a tetrahedral arrangement
c.
four electron pairs in a tetrahedral arrangement
a.
trigonal pyramidal (there is a lone pair on N)
b.
nonlinear, V-shaped (four electron pairs on Se, but only two atoms are attached to Se)
c.
tetrahedral (four electron pairs on Si and four atoms attached)
a.
tetrahedral (four electron pairs on C and four atoms attached)
b.
nonlinear, V-shaped (four electron pairs on S, but only two atoms attached)
c.
tetrahedral (four electron pairs on Ge and four atoms attached)
a.
tetrahedral
b.
tetrahedral
c.
tetrahedral
a.
basically tetrahedral around the P atom (The hydrogen atoms are attached to two of the
oxygen atoms and do not affect greatly the geometrical arrangement of the oxygen atoms
around the phosphorus.)
b.
tetrahedral (4 electron pairs on Cl and 4 atoms attached)
c.
trigonal pyramidal (4 electron pairs on S and 3 atoms attached)
81.
82.
8].
84.
Copyright C Houghtori Mifflin Company. All rgiits reserved.
262
Chapter 12: ChemIcal Bonding
85.
a.
<109.5°
b.
<109.5°
c.
109.5°
d.
109.5°
a.
approximately 109.5° (the molecule
b.
approximately 109.5° (the molecule is trigonal pyramidal)
c.
109.5°
d.
approximately 120° (the double bond makes the molecule flat)
86.
is
V-shaped or nonlinear)
87.
There would be four electron pairs around the sulfur atom, so the arrangement of the electron
pairs connecting the sulfur atom to the four oxygen atoms would basically be tetrahedral.
Assuming the influence of the hydrogen atoms can be neglected, the molecule overall would give
the impression of being basically tetrahedral,
88.
The ethylene molecule contains a double bond between the carbon atoms. This makes the
molecule planar (flat) with H—C—H and H—C—C bond angles of approximately 120°. The 1,2dibromoethane molecule would not be planar, however. Each carbon would have four bonding
pairs of electrons around it, and so the orientation around each carbon atom would be basically
tetrahedral with bond angles of approximately 109.5° (assuming all bonds are similar).
89.
Resonance is said to exist when more than one valid Lewis structure can be drawn for a molecule.
The actual bonding and structure of such molecules is thought to be somewhere “in between” the
various possible Lewis structures. in such molecules, certain electrons are delocalized over more
than one bond.
90.
double
91.
repulsion
92.
The bond with the larger electronegativity difference will be the more polar bond. See Figure 12.3
for electronegativities.
93.
a.
SF
b.
P-O
c.
C-H
The bond with the larger electronegativity difference will be the more polar bond. See Figure 12.3
for electronegativities.
a,
Br—F
b.
As-O
c.
Pb—C
94.
The bond energy of a chemical bond is the quantity of energy required to break the bond and
separate the atoms.
95,
covalent
Copyright © 1-laughton Mifflin Cornpan. All rights reserved
Chapter 12: ChemIcal Bonding
96.
97.
9$.
9Q.
263
In each ease, the element higher up within a group on the periodic table has the higher
electronegati vily.
a.
13e
b.
N
c.
F
For a bond to he polar covalent, the atoms involved in the bond must have different
eleetronegativities (must be oF di I’Ferent elements).
a.
polar covalent
b.
covalent
c.
covalent
d.
polar covalent
For a bond to be polar covalent, the atoms involved in the bond must have different
electronegativities (must be of different elements).
a.
polar covalent (different elements)
h.
nonpolar covalent (two atoms of the same element)
c.
polar coa1ent (different elements)
d.
nonpolar covalent (atoms of the same element)
Electronegativity differences are given in parentheses.
a.
N—P (0.9); N—O (0,5); the N—P bond is more polar.
b.
N—C (0.5): N—O (0.5); the bonds are of the same polarity.
c.
N—S (0.5); N—C (0.5); the bonds are of the same polarity.
d.
N—F 10): N—S (0.5); the N—F bond is more polar.
100. In a diatomic molecule containing two different elements, the more electronegative atom will be
the negative end of the molecule, and the less electronegative atom will be the positive end.
a.
oxygen
b.
bromine
c.
iodine
101. in the figures, the arrow points toward the more electronegative atom.
a.
:•
N—Cl: T’-.e atoms have ry :meariy’ the same eectronegativity, so there s a very small, if
any, dpo:e momer.t.
P—N&-
c.
d.
6--C--\8--
-c’ c.
C.y. ny. A
eevet
264
Chapter 12: Chemical Bonding
:
..-
102.
a.
Al 2
s 2s’ 2p
1
6 32 3p
a
15
6
2 2p
2s
3
AI
Ne has the same configuration as Al
.
3
b.
Br 1s2 2? 21,6
32
6
3p
42
5
3d° 4p
Br Is’ 2? 2p
6 3? 3? 4? 3d” 4,,6
Kr has the same configuration as Br
c.
2 2s’ 2’ 32 3? 4?
Ca is
Ca’ Is
2 2s’ 2? 3? 3?
Ar has the same configuration as (a’.
d.
Li Is’ 2?
•-*
Li is
1
2
.2
He has the same configuration as IJ.
e.
1
2 2? 2p
F Is
F Is’2s’2p
6
Ne has the same configuration as F
....:::
103.
a.
Na
b.
C.
‘I
d,
2
S
f.
Mg’
g.
3
AI
h.
1
N
a.
Na,Se: Na has one electron more than a noble gas: Sc has two fewer electrons than a noble
gas.
b.
RbF: Rb has one electron more than a noble gas, F has one electron less than a noble gas.
c.
K,Te: K has one electron more than a noble gas; Te has two fewer electrons than a noble
gas.
d.
BaSe: Ba has two electrons more than a noble gas; Se has two fewer electrons than a noble
gas.
e.
KAt: K has one ekctron more than a noble gas; At has one electron less than a noble gas.
f.
FrCl: Fr has one electron more than a noble gas; CI has one electron less than a noble gas.
I 04.
Copyright
Hougnion MiItHn Company All rights reserved
-
Chapter 12: Chemical Bonding
105.
a.
2 [Ar]; Br [Kj
Ca
b.
3
A1
c.
2 [Kr]; O2 [Ne]
Sr
d.
2 [Ar]
4 [Ar]; S
K
[Kr]
2
[Nc]; Se
106. Relative ionic sizes are indicated in Figure 12.9.
a.
Na
b.
3
A1
c.
F
d.
Na
a.
He
107,
He:
b.
Br
Br
c.
Sr
Sr:
d.
Ne
:Ne:
e.
I
f.
Ra
Ra:
108.
a.
b.
c.
d.
H provides I; N provides 5; each 0 provides 6; total valence electrons = 24
each H provides 1; S pro’ides 6; each 0 provides 6; total valence electrons = 32
each H provides 1; P provides 5; each 0 provides 6; total valence electrons = 32
H provides 1; Cl provides 7; each 0 provides 6; total valence electrons = 32
Coig © HoLghtor Mifn Corrpany. AlL rights reserved.
.
265
266
chapter 12: ChemIcal Bonding
109.
a.
GeH. Ge provides 4 valence electrons. Each H provides I valence electron. Total valence
:
4
electrons = 8
H
H—Ge—Il
b.
H
ICI: I provides 7 valence electrons. Cl provides 7 valence electrons Total valence electrons
= 14
:k ç,l:
c.
NI N provides 5 valence electrons. Each I provides 7 valence electrons. Total valence
:
1
electrons = 26
:1—N —I:
::
d.
: P provides 5 valence electrons. Each F provides 7 valence electrons. Total valence
3
PF
electrons 26
F—P—F:
F:
110.
a.
N Each N provides 5 valence electrons. Each H provides I valence electron. Total
:
4
H
2
valence electrons = 14
H—N—N—Il
I
b.
I
HH
: Each C provides 4 valence electrons. Each H provides 1 valence electron. Total
6
H
2
C
valence electrons = 14
HH
I
I
H—C
c.
—
I
I,
C
—
H
11 H
: N provides 5 valence electrons. Each Cl provides 7 valence electrons. Total valence
3
NCI
electrons = 26
Cl—N—Cl:
CopyrigP @ Haughton Mi1fIn Conipany AN rights reserved
Chapter 12: ChemIcal Bonding
d.
267
SiCL: Si provides 4 valence electrons. Each Cl provides 7 valence electrons. Total valence
electrons = 32
Cl
Cl—Si—Cl
:Cl:
ill.
a.
SO: S provides 6 valence electrons. Bach 0 provides 6 valence electrons. Total valence
electrons 18
0S—O:
b.
:0—S(s)
4
: Each N provides 5 valence electrons. 0 provides 6 valence electrons. Total valence
N
0
2
electrons 16
:NN—0:
*
c.
03: Each 0 provides 6 valence electrons. Total valence electrons
0=0—0
18
Q—00
112
a
N0 N provides 5 valence electrons Each 0 provides 6 valence electrons The 1— charge
3
means one additional valence electron. Total valence electrons = 24
Q=N—Q:
s
C0 : C provides 4 valence electrons. Each 0 provides 6 valence electrons. The 2— charge
means two additional valence electrons. Total valence electrons 24
0C—O:
I
—
:0:
c.
II
:0:
:0:
b.
0—N=ö
“I..
I
.
:o—c=ö
I
..
:0:
.
:Q——Q:
II
:0:
NK N provides 5 valence electrons. Each H provides I valence electron. The 1+ charge
:
4
means one less valence electron. Total valence electrons = 8
H
H-N-H
[1
, there are single bonds between the
2
113. Beryllium atoms only have two valence electrons. In BeE
2 thus has two electron pairs
beryllium atom and each fluorine atom. The beryllium atom of BeF
in addition to the bonding
molecule,
the
water
another.
For
one
around it that lie 180° apart from
there
are two nonbonding
atoms,
pairs of electrons that attach the hydrogen atoms to the oxygen
pairs of electrons that affect the [1—0—H bond angle.
Couright © Houghion Mifflin Company. All righfs reserved.
268
Chapter 12: Chemical Bonding
114.
a.
four electron pairs arranged tetrahedrally about C
b.
four electron pairs arranged tetrahedrally about Ge
c.
three electron pairs arranged trigonally (planar) around B
a.
nonlinear (V-shaped due to lone pairs on 0)
b.
nonlinear (V—shaped due to lone pairs on 0)
c.
tetrahedral
a.
. trigoiial pyramid (lone pair on Cl)
3
CIO
1,.
, nonlinear (V—shaped. two lone pairs on Ci)
2
ClO
c.
ClOJ, tetrahedral (all pairs on Cl are bonding)
a.
<
b.
109.5° (molecule is tetrahedral)
c.
1 80° (molecule is linear)
d.
120° (molecule is trigonal planar)
a.
nonlinear (V—shaped)
b.
trigonal planar
c.
basically trigonal p!anar around tbe C’ (tl’e
the shape around the carbon only slightly)
d.
linear
a.
trigonal planar
b.
basically trigonal planar around the N (the I--I is attached to
the shape around the nitrogen only slightly)
c.
nonlinear (V—shaped)
d.
linear
115.
I 16.
117.
109.5° (molecule is nonlinear, V-shaped)
118.
attached to one ni the 0 atoms and distorts
119.
one
of the C) atoms and distorts
120. Ionic compounds tend to be hard. crystalline substances with relatively high melting and boiling
points. Covalently bonded substances tend to he gases, liquids, or relatively soil solids with much
lower melting and boiling points.
12 I In a covalent bond between two atoms of the same element, the electron pair is shared equally and
the bond is nonpolar w’ith a bond bet ecu atoms of different elements, the electron pair is
unequally shared, and the bond is polar (assuming the elements have different electronegatiities).
.
pv
ese-vec